Symmetric basis of Banach space ( A Uniformly bounded family) Generally, we have the following definition: Let X be a Banach space with $(e_n)$ basis of X. Then $(e_n)$ called symmetric basis, if for every bijection $\pi\colon N\to N$  the sequence $(e_{\pi(n)})$ is equivalent to the basis $(e_n)$.
In Lindenstrauss-Tzafriri's book "Classical Banach Spaces I" there is a part I can not understand. In the image that I inserted you can see the proof that the family $\{V_\pi\}_\pi$ is uniformly bounded. So, we argue with contradiction and we suppose that $\{V_\pi\}_\pi$ isn't uniformly bounded and we will contradict the condition that $(e_n)$ is symmetric basis . So, there is exist an $x\in X$ : $$\sup_{\pi} \|V_\pi(x)\|=\sup_{\pi}\left\|\sum\limits_{n=1}^\infty \alpha_n e_\pi(n)\right\|=+\infty.$$ Also, understand the first step of the induction, but then I am losing it. Please, can someone help me to figure out the next part of the proof ? Thanks.



 A: Using the notation from Lindenstrauss and Tzafriri, because I think it will be easier if you can compare it with their book:
Let $\{x_n\}_{n\in{\mathbb N}}$ be a symmetric basis for the Banach space $X$ and define a permutation operator $V_\pi:X \rightarrow X$ such that $V_\pi (\sum_{n\in {\mathbb N}}\alpha_n x_n) = \sum_{n\in {\mathbb N}}\alpha_n x_{\pi(n)} $ where $\pi :{\mathbb N} \rightarrow {\mathbb N}$ is a homeomorphism.  We can see that $V_\pi$ is an automorphism of $X$ since $\{x_n\}_{\mathbb N}$ is a symmetric basis and $\pi$ is a homeomorphism.
By the Uniform Boundedness Theorem we know that $\|V_\pi \|$ iff $\sup_\pi \| V_\pi x\|$ is bounded for all $x\in X$, so if $\| V_\pi\|$ is unbounded then there must be some $x$ for which $\|V_\pi x \|$ is unbounded, and in particular there must be some permutation $\pi$ for which this is true as well.
Now, $\pi$ is a permutation so we can decompose it into a finite collection of disjoint cycles (see later for exactly why this is true), so that $\pi = \pi_1 \cdot \pi_2 \cdot \pi_3 \cdot \cdots \pi_k$ where each $\pi_i$ operates on a subset $\sigma _i$ of ${\mathbb N}$.  Because these are disjoint we have $\sigma _i \cap \sigma _j = \emptyset$ for $i\not= j$, and likewise $\pi_i(\sigma _i) \cap \pi_j (\sigma _i) = \emptyset$.  So now we can rewrite the unboundedness condition as
$$ \left\| \sum_{n\in \sigma _j} \alpha _n x_{\pi_j(n)} \right\| \geq 1 $$
(In fact, we can replace the $1$ by $\varepsilon$ -- the key thing is that it's a quantity bounded away from $0$ so that the infinite sum diverges).
Now define a new permutation $\pi _0$ by $\pi_0 (n) = \pi_{2j}(n)$ for $n \in \sigma_{2j}$.  We claim that the permutation $\sum_{n\in {\mathbb N}} \alpha_n x_{\pi(n)}$ diverges, which contradicts $\{x_n\}_{n\in {\mathbb N}}$ being symmetric.
$\sum_{n\in {\mathbb N}} \alpha_n x_{n}$ converges if, for every $\varepsilon >0$ there exists a finite set $\sigma $ such that for every finite set $\tau \supset \sigma$ we have $\|\sum_{n \in \tau} \alpha_n x_n -x \| < \varepsilon$.  Taking any finite set $\tau$ like this allows us to decompose our permutatation $\pi$ as claimed above, and by the unboundedness condition we see that our new permutation $\pi_0$ will always have infinitely many elements that are bounded away from $0$, which is why our claim for divergence holds.
Ultimately, this completes the proof by contradicting the symmetry of the basis.
Note: page 960 of http://www.ams.org/journals/bull/1940-46-12/S0002-9904-1940-07344-6/S0002-9904-1940-07344-6.pdf (a 4-page paper) lays out the basic argument for the last claim very nicely, as an additional perspective on this may be useful.
